An object is placed 10 cm from a diverging lens, which forms an image 6.5 cm from the lens. What is the focal length of the lens?

-18.5 cm

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+ 3.9 cm

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- 3.9 cm

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+18.5 cm

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Solution

The correct option is A
-18.5 cm

Lens formula , $\frac{1}{v} = \frac{1}{f} + \frac{1}{u}$
$- \frac{1}{6.5} = \frac{1}{f} - \frac{1}{10} \frac{1}{f} = - \frac{1}{6.5} + \frac{1}{10} \frac{1}{f} = \frac{(- 10 + 6.5)}{65} f = - \frac{65}{3.5} = - 18.5 c m$

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An object is placed 10 cm from a diverging lens, which forms an image 6.5 cm from the lens. What is the focal length of the lens?

The correct option is A -18.5 cm Lens formula ,1v=1f+1u −16.5=1f−1101f=−16.5+1101f=(−10+6.5)65f=−653.5=−18.5cm

The correct option is A
-18.5 cm

Lens formula , $\frac{1}{v} = \frac{1}{f} + \frac{1}{u}$
$- \frac{1}{6.5} = \frac{1}{f} - \frac{1}{10} \frac{1}{f} = - \frac{1}{6.5} + \frac{1}{10} \frac{1}{f} = \frac{(- 10 + 6.5)}{65} f = - \frac{65}{3.5} = - 18.5 c m$